To avoid confusion with other uses of braces, let $F:\Bbb R\to[0,1)$ be the fractional part function (usually noted as $\{\cdot\}$), so $F(x)=x-\lfloor x\rfloor$.
It is known that the set $S:=\{F(\sqrt n):n\in\Bbb N\}$ is dense in $[0,1]$, because $S$ contains, for example, $\{F(n\sqrt 2):n\in\Bbb N\}$.
But is the set $$\{F(\sqrt p):p\text{ prime}\}$$ also dense in $[0,1)$? I had been thinking on it, and my intuition says that it is, but I'm clueless about how to prove it. (In fact, I'm not sure even about the most appropriate tags for this question).
For $\Re(z) > 0$ let $$g(z)=\sum_p e^{-p^{1/2}z} = \sum_{n \ge 2} \pi(n)(e^{-n^{1/2}z}-e^{-(n+1)^{1/2}z})=\sum_{n \ge 2} \frac{n}{\ln n}(1+o(1))e^{-n^{1/2}z} n^{-1/2}(1+o(1)) $$
The point is that for any fixed $y \ne 0$ then $$|g(x+2i\pi y)|\le g(x) \qquad and \qquad g(x+2i\pi y) = o(g(x)) \qquad as\quad x\to 0^+$$
For any $\phi(y) = \sum_k \hat{\phi}(k)e^{2i\pi k y}$ smooth $1$-periodic then
$$\boxed{\sum_p e^{-p^{1/2}x} \phi(p^{1/2})= \sum_k g(x-2i\pi k)\hat{\phi}(k) = \hat{\phi}(0) g(x)+o(g(x))}$$